Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be
$10 cm$ per sec
$2.5 cm$ per sec
$5 \times {(4)^{1/3}}cm$ per sec
$5 \times \sqrt 2 \,cm$ per sec
Two spheres $P$ and $Q$ of equal radii have densities $\rho_1$ and $\rho_2$, respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities $\sigma_1$ and $\sigma_2$ and viscosities $\eta_1$ and $\eta_2$, respectively. They float in equilibrium with the sphere $P$ in $L_1$ and sphere $Q$ in $L _2$ and the string being taut (see figure). If sphere $P$ alone in $L _2$ has terminal velocity $\overrightarrow{ V }_{ P }$ and $Q$ alone in $L _1$ has terminal velocity $\overrightarrow{ V }_{ Q }$, then
$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$
When a body falls in air, the resistance of air depends to a great extent on the shape of the body, $ 3 $ different shapes are given. Identify the combination of air resistances which truly represents the physical situation. (The cross sectional areas are the same).
A small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta $ , then the sphere encounters an opposing force of
A plastic circular disc of radius $R$ is placed on a thin oil film, spread over a flat horizontal surface. The torque required to spin the disc about its central vertical axis with a constant angular velocity is proportional to
Write the equation of terminal velocity.